On the exact separation of mixed integer knapsack cuts - Marcos ...

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26 R. Fukasawa, M. Goycoolea(b) There exist i, j ∈ 1,...,n such that e i > e j ,(a i > 0, c i > 0, u i =+∞) and (a j < 0, c j ≤ 0, u j =+∞).Proof It is clear that if either condition holds then the problem must be unbounded. Weprove the converse. Let U ={i ∈ 1,...,n : u i =+∞}, and consider the recessioncone of conv(K ), which is given by C ={x ∈ R n : ax ≤ 0, x i ≥ 0 ∀i ∈ U, x i =0 ∀i /∈ U}.The extreme rays of conv(K ) are the extreme rays of C and must satisfy n − 1ofthe linearly independent constraints defining C at equality. Thus, all extreme ray vectorshave at most two-nonzero values. Let x correspond to an extreme ray satisfyingcx > 0. We may assume that (a) does not hold. Therefore, for any nonzero componentx i we have that c i > 0 ⇒ a i > 0 and a i ≤ 0 ⇒ c i ≤ 0. Given that cx > 0wemayassume that c i > 0. Thus a i > 0 and since ax ≤ 0, it follows that there must exist jsuch that x j > 0 and a j < 0. Finally, cx > 0 implies that c i x i > −c j x j , and ax = 0implies that a i x i =−a j x j . From this we have that e i > e j , and this concludes ourproof.⊓⊔Observe that Proposition 3 implies that it can be determined if MIKP is unboundedin linear time. In fact, this can be achieved by looping through the unbounded variablesand keeping track of the most efficient one satisfying (a i > 0, c i > 0), the least efficientone satisfying (a j < 0, c j ≤ 0), and checking if any satisfy (a i ≤ 0, c i > 0).3.2 PreprocessingWe utilize a simple eight-step pre-processing procedure in order to speed our algorithm.Let N + ={i ∈ 1,...,n : a i > 0}, and N − ={i ∈ 1,...,n : a i < 0}. Weassume that for each i ∈ I we have l i ∈ Z and u i ∈ Z ∪{∞}. If there exists i ∈ N −such that u i =+∞, define L min =−∞. Otherwise, define,L min = ∑a i u i + ∑a i l ii∈N − i∈N +If there exists i ∈ N + such that u i =+∞, define L max =+∞. Otherwise,L max = ∑a i l i + ∑a i u ii∈N − i∈N +Finally, let sign(x) = 1ifx > 0, sign(x) =−1ifx < 0 and sign(x) = 0ifx = 0.1. Detect infeasibility. The problem is infeasible if and only if one of the followingconditions hold: (a) L min ̸= −∞and L min > b, or (b) there exists i ∈ 1,...,nsuch that u i ̸= ∞and l i > u i .2. Detect unboundedness. Use the algorithm discussed in Sect. 3.1 to detect if theproblem is unbounded. Observe that if the problem is bounded, all variables x isatisfying c i ≥ 0 and a i ≤ 0 will be such that u i is finite.123
On the exact separation of MIK cuts 273. Fix variables at their bounds. Fix to u i all variables x i such that c i ≥ 0 anda i ≤ 0. Fix to l i to all variables x i such that c i ≤ 0 and a i ≥ 0. Observe that afterfixing, all variables will be such that (a i > 0 and c i > 0) or (a i < 0 and c i < 0).4. Tighten upper bounds. If L min ̸= −∞, then the for each i ∈ N + we can re-defineu i := min{u i , (b − L min + a i l i )/a i }.Ifi ∈ I we can further strengthen this byletting u i := ⌊u i ⌋.5. Tighten lower bounds. If L min ̸= ∞, then the for each i ∈ N − we can re-definel i := max{l i , (b − L min + a i u i )/a i }.Ifi ∈ I we can further strengthen this byletting l i := ⌈l i ⌉.6. Sort variables Sort variables in order of increasing efficiency. Break ties by lettinginteger variables precede continuous variables. Break second ties in order ofincreasing a i .7. Aggregate integer variables. If two integer variables x i , x j are such that a i = a jand c i = c j aggregate them into a new variable x k of the same type such thata k = a i = a j , c k = c i = c j , l k = l i + l j and u k = u i + u j .8. Aggregate continuous variables. If two continuous variables x i , x j are such thatand sign(a i ) = sign(a j ) aggregate them into a new variable x k of thec ia i= c ja jsame type such that a k = a i , c k = c i , l k = l i + a ja il j and u k = u i + a ja iu j .Note that steps one through five can be performed together in a single loop, andthat step seven can be performed in a single pass after sorting. For more informationand for possible extensions see Savelsbergh [45]. From this point on, we will assumethat we are always dealing with instances that have been preprocessed according tothe steps above.3.3 Branch and boundWe use a depth-first-search branch and bound algorithm which always branches on theunique fractional variable. We use a simple linear programming algorithm, a variationof Dantzig’s algorithm [20], which runs in linear time by taking advantage of the factthat variables are sorted by decreasing efficiency. We do not use any cutting planesin the algorithm, nor any heuristics to generate feasible solutions. The algorithm usesvariable reduced-cost information to improve variable bounds at each node of the tree.3.4 DominationConsider x 1 and x 2 , two feasible solutions of MIKP. We say that x 1 cost-dominatesx 2 if cx 1 > cx 2 and ax 1 ≤ ax 2 . On the other hand, we say that x 1 lexicographicallydominatesx 2 if cx 1 = cx 2 , ax 1 ≤ ax 2 , and if in addition there exists i ∈{1,...,n}such that xi 1 < xi 2 and xk 1 = x2 kfor all k < i. We say that a solution is dominated if itis cost-dominated or lexicographically-dominated. Observe that there exists a uniquenon-dominated optimal solution (or none at all).Traditional branch and bound algorithms work by pruning nodes when (a) they areproven infeasible, or (b) when the LP relaxation of the nodes have value worse than123
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